Q.If \(a, b, c, d\) are consecutive terms in geometric progression, prove that \((a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2\).

Let, \(\cfrac{a}{b}=\cfrac{b}{c}=\cfrac{c}{d}=k\) [\(k\) is a nonzero constant]
\(\therefore a=bk, b=ck, c=dk\)
\(\therefore b=dk\times k=dk^2, a=dK^2\times k =dk^3\)

LHS \(=(a^2-b^2 )(c^2-d^2 )\)
\(=\{(dk^3)^2-(dk^2)^2\}\{(dk)^2-d^2\}\)
\(=\{d^2k^6-d^2k^4\}\{d^2k^2-d^2\}\)
\(=d^2k^4(k^2-1)\times d^2(k^2-1)\)
\(=d^4k^4(k^2-1)^2\)

RHS \(=(b^2-c^2)^2\)
\(=\{(dk^2)^2-(dk)^2\}^2\)
\(=\{d^2k^4-d^2d^2\}^2\)
\(=\{d^2k^2(k^2-1)\}^2\)
\(=d^4k^4(k^2-1)^2\)

\(\therefore\) LHS = RHS [Proved]
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